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Current Electricity

Question

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In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 ohm is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.

Solution

Potentiometer at open circuit ℓ₁ = 350

R = 9

ℓ₂ = 300

$r = R (\frac{I_{1}}{l_{2}} - 1) r = 9 (\frac{350}{300} - 1) = 1.5 Ω$

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